Q. 114.1( 52 Votes )

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Answer :

Given A and B are symmetric matrix of same order

A = A’ eqn1

B = B’ eqn2

So, AB – BA = A’B’ – B’A’ (from 1 & 2)

AB –BA = (BA)’ – (AB)’ ( )

AB – BA = (-1) ((AB)’ – (BA)’) (taking -1 common)

AB – BA = -(AB – BA)’ ( )

Here we see that the relation between (AB – BA) and its transpose i.e. (AB – BA)’ is (AB –BA) = -(AB – BA)’, this implies that (AB – BA is a skew symmetric matrix.

Hence proved.

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